NCERT Solutions for Class 11 Chemistry Chapter 9: Hydrocarbons

Concept. Chlorination of methane is a free-radical chain reaction with three stages: initiation, propagation and termination. Ethane is not formed by the main propagation steps but by one of the possible termination steps, in which two methyl free radicals combine; its presence is itself good evidence that methyl radicals exist as intermediates.

1. Initiation. In ultraviolet light or at 573–773 K the weakest bond, Cl–Cl, undergoes homolysis: Cl–Cl → Cl• + Cl•. (The Cl–Cl bond is weaker than C–C and C–H bonds, so it breaks first.)
2. Propagation (a). A chlorine radical removes a hydrogen atom from methane: CH₄ + Cl• → •CH₃ + H–Cl. A methyl free radical is formed.
3. Propagation (b). The methyl radical attacks a chlorine molecule: •CH₃ + Cl–Cl → CH₃–Cl + Cl•. Chloromethane is the principal product and a new Cl• is regenerated, so steps (a) and (b) repeat as a chain.
4. Termination. The chain stops when radicals meet and pair up their odd electrons. Three possibilities exist: (i) Cl• + Cl• → Cl–Cl; (ii) H₃C• + •CH₃ → CH₃–CH₃ (ethane); (iii) •CH₃ + Cl• → CH₃–Cl.

Why only a trace of ethane? Radicals are present in very low concentration, so the chance that two methyl radicals meet is far smaller than the chance that a methyl radical meets one of the abundant Cl₂ molecules. Termination therefore competes only weakly with propagation, and ethane is a by-product. Once ethane is formed it can itself be chlorinated to chloroethane (C₂H₆ + Cl• → •C₂H₅ + HCl; •C₂H₅ + Cl₂ → C₂H₅Cl + Cl•), which is why small amounts of C2 products accompany the chloromethanes.

Extra depth. Further chlorination. CH₃Cl is attacked in the same way (CH₃Cl + Cl• → •CH₂Cl + HCl; •CH₂Cl + Cl₂ → CH₂Cl₂ + Cl•), giving CH₂Cl₂, CHCl₃ and CCl₄ in turn. To favour CH₃Cl a large excess of methane is used; to reach CCl₄ an excess of chlorine is used. One photon can start a chain of thousands of propagation steps, so the process is a true chain reaction. Reaction in the dark at room temperature does not occur because Cl• is not formed.

Conclusion: Ethane is formed in the termination stage by the coupling of two methyl free radicals, CH₃• + •CH₃ → CH₃–CH₃.

Concept. Follow the IUPAC sequence: (1) find the longest chain containing the principal groups (multiple bonds are included even if a longer saturated chain exists); (2) number to give the lowest locants to multiple bonds (a double bond gets preference over a triple bond only when there is a choice at the same locant); (3) name and number the substituents; (4) arrange alphabetically; (5) add suffix -ene / -yne / -diene.

Check (g) carbon by carbon (from the CH₂= end): C1 (CH₂=), C2 (=CH–), C3 (CH₂), C4 (CH with ethyl), C5 (=CH–), C6 (=CH–), C7 (CH₂), C8 (=CH–), C9 (=CH–), C₁₀ (CH₃). Double bonds C1=C2, C5=C6, C8=C9, so the locants are 1, 5, 8.

Extra depth. In (f) students often choose the chain through the branch because it looks ‘straight’; always count carbon atoms, not the picture. In (a) the C=C gets locant 2 from either end, so the tie is broken by the substituent: the CH₃ gets locant 2 (name 2-methylbut-2-ene, not 3-methylbut-2-ene). For polyenes the name keeps the ‘a’ (buta-, deca-) before di/tri/tetra.

Conclusion: Names: (a) 2-methylbut-2-ene, (b) pent-1-en-3-yne, (c) buta-1,3-diene, (d) 4-phenylbut-1-ene, (e) 2-methylphenol, (f) 5-(2-methylpropyl)decane, (g) 4-ethyldeca-1,5,8-triene.

Concept. Both formulae fit C_nH_2n (alkene) and C_nH_2n−2 (alkyne). Fix the carbon skeleton first (continuous chain or branched), then move the multiple bond to every non-equivalent position, and finally check for geometrical (cis–trans) isomers.

(a) C₄H₈: one C=C

1. Straight chain, double bond at the end: CH₂=CH–CH₂–CH₃ — but-1-ene.
2. Straight chain, double bond in the middle: CH₃–CH=CH–CH₃ — but-2-ene. It exists as two geometrical isomers: cis-but-2-ene and trans-but-2-ene.
3. Branched chain: CH₂=C(CH₃)–CH₃ — 2-methylprop-1-ene (isobutylene).

(b) C₅H₈: one C≡C

1. Straight chain: HC≡C–CH₂–CH₂–CH₃ — pent-1-yne; CH₃–C≡C–CH₂–CH₃ — pent-2-yne.
2. Branched chain (four-carbon chain with a methyl): HC≡C–CH(CH₃)–CH₃ — 3-methylbut-1-yne. (A methyl on C-2 of a butyne is impossible because a triply bonded carbon already has four bonds.)

Alkynes show no geometrical isomerism because the C≡C unit and its two attached atoms lie in a straight line, so there are no ‘two sides’ for groups to occupy.

Extra depth. The number of alkene isomers of C₄H₈ is 3 structural (4 counting the cis/trans pair), exactly as in the NCERT text; cyclobutane and methylcyclopropane are also C₄H₈ but contain no double bond, so they are excluded. For C₅H₈ the three alkyne isomers are all that exist; the total number of C₅H₈ isomers is larger only if rings and dienes are allowed.

Conclusion: (a) But-1-ene, but-2-ene (cis and trans) and 2-methylprop-1-ene. (b) Pent-1-yne, pent-2-yne and 3-methylbut-1-yne.

Concept. Ozonolysis cleaves the C=C bond: O₃ adds to give an ozonide, and Zn/H₂O decomposes it into two carbonyl compounds. Rule: split the molecule at the double bond and put =O on each carbon that was doubly bonded. A carbon carrying an H becomes an aldehyde (–CHO); a carbon carrying two carbon groups becomes a ketone (C=O); a =CH₂ becomes methanal (HCHO).

Writing the structure of (ii): hept-3-ene means CH₃–CH₂–CH=CH–CH₂–CH₂–CH₃; put CH₃ on C-3 and C-4. C-3 then carries ethyl + methyl and C-4 carries methyl + propyl, so both fragments are ketones: the left fragment has 4 carbons (butan-2-one) and the right fragment has 5 carbons (pentan-2-one).

Extra depth. Ozonolysis is used to locate the double bond: from the carbon counts of the two fragments one can rebuild the alkene (see Ex 9.5–9.7). It cannot distinguish cis from trans because the original geometry is destroyed. In water-free work-up without Zn, the aldehydes would be oxidised by H₂O₂ to carboxylic acids, so Zn (or dimethyl sulphide) is used to keep aldehydes.

Conclusion: (i) Ethanal + propanal; (ii) butan-2-one + pentan-2-one; (iii) methanal + pentan-3-one; (iv) benzaldehyde + propanal.

Concept. Ozonolysis is reversed by joining the two carbonyl carbons through a double bond: remove the oxygen atoms from both products and connect the carbons that carried them by C=C. Ethanal supplies one half, pentan-3-one the other.

1. Write the two products: CH₃–CH=O (ethanal) and CH₃CH₂–C(=O)–CH₂CH₃ (pentan-3-one).
2. Delete both oxygens: CH₃–CH< and >C(CH₂CH₃)₂.
3. Join the two free valences with a double bond: CH₃–CH=C(CH₂CH₃)₂, i.e. (C₂H₅)₂C=CH–CH₃.
4. Name it: longest chain containing C=C is C-C-C-C-C (five carbons): CH₃(1)–CH(2)=C(3)(C₂H₅)–CH₂(4)–CH₃(5). Double bond at 2, ethyl on C-3.

Check of the molecular formula: alkene C₇H₁₄ (C_nH_2n, n = 7). The two products together have 2 + 5 = 7 carbons, matching.

Extra depth. The alkene shows geometrical isomerism? One double-bonded carbon (C-3) carries two identical ethyl groups, so there is no cis–trans isomerism; the structure is unique. If the question had said that A gave ethanal and propanone the answer would be 2-methylbut-2-ene, so the products must always be matched carbon-for-carbon.

Conclusion: A is CH₃–CH=C(C₂H₅)₂, named 3-ethylpent-2-ene.

Concept. Use the bond count to find the formula and skeleton, then use the molar mass of the ozonolysis product to find the symmetry of the alkene. Remember that a C=C contributes one σ and one π bond, so the three C–C σ bonds include the σ part of the double bond.

1. Number of carbons. In an acyclic skeleton, n carbons are joined by (n − 1) C–C σ bonds. Three C–C σ bonds ⇒ 4 carbons.
2. Number of hydrogens. Eight C–H σ bonds ⇒ 8 hydrogens. Formula = C₄H₈, which fits C_nH_2n with one C=C (one π bond) ✓.
3. Which aldehyde has M = 44 u? RCHO: M(CHO) = 29, so R = 15 = CH₃. The aldehyde is ethanal, CH₃CHO (2C + 4H + O = 24 + 4 + 16 = 44 u).
4. Two moles of the same aldehyde from one mole of alkene means the molecule is symmetrical about C=C: CH₃–CH=CH–CH₃.
5. Check: CH₃CH=CHCH₃ → (O₃, Zn/H₂O) → 2 CH₃CHO ✓.

Bond count of but-2-ene: C–C σ = 3 (C1–C2, C2–C3, C3–C4); C–H σ = 3 + 1 + 1 + 3 = 8; C–C π = 1. All three conditions are satisfied.

Extra depth. Both the cis and trans forms of but-2-ene fit the data; the question asks only for the IUPAC name, so but-2-ene is complete. If the aldehyde had M = 58 u the symmetrical alkene would be hex-3-ene (2 propanal), and for M = 30 u (methanal) the alkene would be ethene; this molar-mass method fixes the aldehyde quickly.

More to know — why ozonolysis is useful. Before spectroscopy, ozonolysis was the standard way to prove where the double bond lies: the carbon counts of the two fragments locate it. Ozonolysis of cyclic alkenes (e.g. cyclohexene) gives a single dicarbonyl compound (hexanedial) because both ends of the cleaved bond stay in one molecule, while a diene gives three fragments (buta-1,3-diene gives 2 methanal + ethanedial).

Conclusion: A is but-2-ene, CH₃–CH=CH–CH₃.

Concept. The alkene is obtained by removing the oxygen atoms of the two carbonyl compounds and joining their carbonyl carbons with a double bond (the reverse of ozonolysis).

1. Propanal: CH₃CH₂–CH=O → fragment CH₃CH₂–CH<.
2. Pentan-3-one: CH₃CH₂–C(=O)–CH₂CH₃ → fragment >C(CH₂CH₃)₂.
3. Join: CH₃CH₂–CH=C(CH₂CH₃)₂, i.e. (C₂H₅)₂C=CH–C₂H₅. Molecular formula C₈H₁₆ (3 + 5 carbons).
4. Name: the longest chain through the double bond is six carbons: CH₃(1)–CH₂(2)–C(3)(C₂H₅)=CH(4)–CH₂(5)–CH₃(6) ⇒ 3-ethylhex-3-ene.

Forward check. 3-Ethylhex-3-ene: C-3 carries two ethyl groups (the CH₂CH₃ of the chain and the ethyl branch) and C-4 carries one H and an ethyl group. On ozonolysis C-3 becomes C=O with two ethyl groups (pentan-3-one) and C-4 becomes CHO with one ethyl group (propanal) ✓.

Extra depth. Compare with Ex 9.5: there the aldehyde was ethanal, giving 3-ethylpent-2-ene; here the aldehyde is propanal, so the chain is one carbon longer: 3-ethylhex-3-ene. The compound has no cis–trans isomers because one doubly bonded carbon has two identical ethyl groups.

More to know — reverse-ozonolysis practice. (1) Two moles of propanal → hex-3-ene, CH₃CH₂CH=CHCH₂CH₃. (2) Methanal + propanone → 2-methylprop-1-ene, CH₂=C(CH₃)₂. (3) Ethanal + butan-2-one → 3-methylpent-2-ene, CH₃CH=C(CH₃)CH₂CH₃. (4) Benzaldehyde + ethanal → 1-phenylprop-1-ene, C₆H₅CH=CHCH₃. In every case the number of carbons of the alkene equals the sum of the carbons of the two carbonyl compounds, a quick check for your answer.

Conclusion: The alkene is CH₃CH₂–CH=C(CH₂CH₃)₂, 3-ethylhex-3-ene.

Concept. Complete combustion of any hydrocarbon in excess oxygen gives CO₂ and H₂O and releases heat. The general balanced equation (NCERT eq. 9.83) is: C_xH_y + (x + y/4) O₂ → x CO₂ + (y/2) H₂O.

Atom balance check (hexyne as an example)

1. Carbon: 6 C on the left, 6 CO₂ on the right ⇒ 6 C ✓.
2. Hydrogen: 10 H on the left, 5 H₂O on the right ⇒ 10 H ✓.
3. Oxygen: right side has 6 × 2 + 5 = 17 O atoms, so 17/2 O₂ molecules are needed on the left ✓.

Energy and flame. All combustion reactions are exothermic (for example CH₄: ΔH = −890 kJ mol⁻¹; butane: −2875.84 kJ mol⁻¹), which is why alkanes (LPG, CNG, petrol) are fuels. Hydrocarbons richer in carbon (alkynes, arenes such as toluene) burn with a luminous, sooty flame because the high C : H ratio leaves unburnt carbon particles; in limited air carbon black (C) and CO are formed instead of CO₂.

Extra depth. In a limited supply of air butane burns incompletely: 2C₄H₁₀ + 9O₂ → 8CO + 10H₂O (carbon monoxide) or gives carbon black; the question asks only for combustion in excess of oxygen. Carbon monoxide is a poison because it binds haemoglobin, so gas burners must be well ventilated.

Conclusion: (i) C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O; (ii) C₅H₁₀ + 15/2 O₂ → 5CO₂ + 5H₂O; (iii) C₆H₁₀ + 17/2 O₂ → 6CO₂ + 5H₂O; (iv) C₇H₈ + 9O₂ → 7CO₂ + 4H₂O.

Concept. Hex-2-ene is CH₃–CH=CH–CH₂CH₂CH₃. Each doubly bonded carbon carries two different groups (H and CH₃ on C-2; H and CH₂CH₂CH₃ on C-3), so geometrical isomerism is possible because rotation about C=C is restricted by the π bond. Cis = the two larger groups (CH₃ and C₃H₇) on the same side; trans = on opposite sides.

Why the cis isomer boils higher. In the cis form the two alkyl groups are on one side, so their small bond moments reinforce each other, giving the molecule a net dipole moment. In the trans form the moments point in opposite directions and cancel (as in trans-but-2-ene, μ ≈ 0, while cis-but-2-ene has μ = 0.33 D). A polar molecule has extra dipole–dipole attraction, so more energy is needed to separate the molecules: higher b.p.

Extra depth. The difference between the boiling points of cis and trans isomers is usually small (a degree or two) because the dipole moments are small; the effect is greater when electronegative substituents are present, e.g. cis-1,2-dichloroethene boils at 60 °C and the trans isomer at 48 °C. For melting points the order is usually reversed: the more symmetrical trans isomer packs better in the crystal and has the higher m.p.

Conclusion: Cis-hex-2-ene has the higher boiling point because it has a net dipole moment, whereas the dipoles in the trans isomer cancel.

Concept. Benzene does not contain three localised C=C bonds. It is a resonance hybrid with six delocalised π electrons spread over six carbon atoms, and this delocalisation lowers its energy far below that of a hypothetical cyclohexa-1,3,5-triene.

1. Structure. All six carbons are sp² hybridised; the six C–C σ bonds and six C–H σ bonds lie in one plane (planar hexagon, 120°). Each carbon keeps one unhybridised p orbital perpendicular to the plane.
2. Delocalisation. Each p orbital overlaps equally with both neighbours, forming two doughnut-shaped π clouds, one above and one below the ring. The six π electrons are not tied to any two carbons.
3. Resonance. The two Kekulé structures A and B are equivalent contributors; the real molecule is their hybrid (drawn with a circle in the hexagon), more stable than either.
4. Evidence from bond lengths. X-ray diffraction shows all six C–C bonds are equal (139 pm), between C–C (154 pm) and C=C (133 pm). In cyclohexatriene they would alternate (154 and 133 pm).
5. Nuclear attraction. A delocalised π cloud is attracted by six nuclei, a localised one by only two, so the delocalised cloud is held more firmly and the molecule is more stable.
6. Consequence. Benzene resists addition (which would destroy the delocalised system) and prefers substitution, which keeps the aromatic ring.

Thermochemical proof. Hydrogenation of one C=C in cyclohexene releases about 120 kJ mol⁻¹; three isolated C=C bonds would release about 360 kJ mol⁻¹. Benzene releases only about 208 kJ mol⁻¹, i.e. it is about 152 kJ mol⁻¹ more stable than the localised-bond model: this energy is the resonance (delocalisation) energy. (These energy values are standard textbook thermochemical data, given here as a supplement to the NCERT explanation.)

Extra depth. Benzene fulfils the three requirements for aromaticity: planar ring, complete cyclic delocalisation and (4n + 2) = 6 π electrons (n = 1). The extra stabilisation of aromatic compounds is called aromatic stabilisation. In the arenium-ion mechanism of substitution the driving force of the last step (loss of H⁺) is regaining this stabilisation.

Conclusion: Benzene is extra-stable because its six π electrons are completely delocalised over the planar ring (resonance), giving equal C–C bonds of 139 pm and about 152 kJ mol⁻¹ of extra stability.

Concept. A cyclic system is aromatic (extra stable, with benzene-like chemistry) if it satisfies three conditions together. The ring does not have to contain a benzene ring; the conditions are about the electrons and the shape.

1. Planarity. The ring must be planar (all ring atoms sp² or otherwise coplanar) so that the p orbitals are parallel and can overlap sideways.
2. Complete delocalisation of π electrons. There must be a continuous, uninterrupted cycle of overlapping p orbitals around the ring: every ring atom must carry a p orbital (no sp³ carbon in the ring).
3. Hückel rule: (4n + 2) π electrons in the ring, where n = 0, 1, 2, 3 … i.e. 2, 6, 10, 14 … π electrons.

How to apply the test (3 questions): (1) Is every ring atom sp² (or carries a p orbital) and is the ring flat? (2) Is the cycle of p orbitals unbroken? (3) Count the π electrons: does the number fit 2, 6, 10, 14? If any answer is ‘no’, the system is not aromatic.

Extra depth. Planar cyclic systems with 4n π electrons (4, 8, 12 …), e.g. cyclobutadiene, are even less stable than expected and are called antiaromatic. Aromaticity is also what makes benzenoid and non-benzenoid compounds (such as the cyclopentadienyl anion or tropylium cation) undergo electrophilic substitution rather than addition. Heterocycles such as pyridine, pyrrole and furan are aromatic as well (Class XII).

Conclusion: A system is aromatic if it is planar, has a complete cyclic delocalisation of π electrons, and contains (4n + 2) π electrons (Hückel rule).

Concept. Test each ring against the three conditions: planarity, uninterrupted cyclic delocalisation, and (4n + 2) π electrons. Failing even one condition is enough to make a ring non-aromatic.

Tip for (i): although the molecule has 6 π electrons, they are not all in one closed loop of p orbitals: the CH₂ in the ring (sp³) breaks the cycle, so the number 6 alone is not enough. This is why the planarity/delocalisation conditions come before counting electrons.

Extra depth. Cyclopentadiene has an interesting property: its CH₂ hydrogens are unusually acidic (pK_a about 16) because loss of H⁺ leaves the cyclopentadienyl anion, which has 6 π electrons in a planar ring and is aromatic. Similarly tropylium cation C₇H₇⁺ (6 π electrons) is aromatic. Cyclooctatetraene is stable enough to be bottled, but it behaves like an ordinary polyene (it adds bromine easily), confirming that it is not aromatic.

Conclusion: None of the three satisfies all three conditions: (i) and (ii) contain an sp³ carbon (no continuous delocalisation, and (ii) has only 4 π electrons) and (iii) has 8 π electrons and is non-planar.

Concept. Plan every conversion with the directive influence of the group already on the ring: groups –Br, –Cl and –CH₃ are ortho/para directing, –NO₂ is meta directing. The order of the two introduction steps therefore decides whether the product is para or meta.

(i) p-Nitrobromobenzene (1-bromo-4-nitrobenzene)

Bromine is o,p-directing, so nitration of bromobenzene gives mainly the para product (para is favoured over ortho for steric reasons); the small amount of ortho isomer is removed by crystallisation or distillation.

(ii) m-Nitrochlorobenzene (1-chloro-3-nitrobenzene)

Nitration must come first: the nitro group is meta-directing, so chlorination of nitrobenzene (a deactivated ring, hence a Lewis-acid catalyst and heat are needed) goes to the meta position.

(iii) p-Nitrotoluene (1-methyl-4-nitrobenzene)

Alkylation comes first because Friedel–Crafts reactions do not work on nitrobenzene (the ring is too deactivated). The methyl group then directs the nitro group to ortho and para; the para isomer is separated.

(iv) Acetophenone (methyl phenyl ketone)

This is Friedel–Crafts acylation: the electrophile is the acylium ion CH₃C⁺=O, formed by CH₃COCl + AlCl₃. Acetic anhydride, (CH₃CO)₂O, with anhydrous AlCl₃ also works.

Extra depth. Mechanism of the nitration step: HNO₃ + H₂SO₄ → NO₂⁺ (nitronium ion) + HSO₄⁻ + H₂O; NO₂⁺ attacks the π cloud to form the σ-complex, which loses H⁺ to give the nitro compound. If the order of steps in (ii) were reversed (chlorinate first), the product would be the ortho/para isomer, not meta: this is the usual ‘order of reactions’ question.

Conclusion: (i) Br₂/FeBr₃ then HNO₃/H₂SO₄; (ii) HNO₃/H₂SO₄ then Cl₂/FeCl₃; (iii) CH₃Cl/AlCl₃ then HNO₃/H₂SO₄; (iv) CH₃COCl/anhydrous AlCl₃.

Concept. A carbon is classified by the number of other carbon atoms attached to it: primary (1°) – one; secondary (2°) – two; tertiary (3°) – three; quaternary (4°) – four. The number of H atoms on it is 4 minus that number (carbon is tetravalent and saturated).

Check: 5 + 2 + 1 + 1 = 9 carbons; hydrogens 15 + 4 + 1 + 0 = 20 ⇒ C₉H₂₀, which satisfies C_nH_2n+2 (2 × 9 + 2 = 20) ✓. IUPAC name of the compound: 2,4,4-trimethylhexane.

Extra depth. The question asks for the 1°, 2°, 3° carbons (the NCERT numbering 1 ,2 ,3 carbon atoms); the quaternary carbon is listed above for completeness. The class of carbon matters in reactivity: free-radical halogenation of C–H bonds follows 3° > 2° > 1° because the radicals formed have the same stability order, and KMnO₄ oxidises only alkanes with a 3° hydrogen (e.g. 2-methylpropane).

Conclusion: There are five 1° carbons (3 H each), two 2° carbons (2 H each) and one 3° carbon (1 H); the quaternary carbon carries no hydrogen.

Concept. Branching lowers the boiling point: among isomeric alkanes, the straight-chain isomer boils highest and the most highly branched isomer boils lowest.

1. Alkanes are non-polar; the only attractions between molecules are weak van der Waals (London dispersion) forces.
2. These forces depend on the surface area of contact between molecules. A straight-chain molecule is long and can lie alongside its neighbours, giving a large contact area.
3. With branching the molecule becomes more compact and, in the extreme, spherical (neopentane). The area of contact decreases, so the intermolecular forces become weaker.
4. Weaker forces are overcome at a lower temperature, so the boiling point falls: 309.1 K > 300.9 K > 282.5 K for the three pentanes.

Related trends. For a straight-chain series the boiling point rises steadily with molecular mass (CH₄ 111 K, C₁₀H₂₂ 447.1 K, C₂₀H₄₂ 615 K) because the surface area and the forces increase. The melting point of branched isomers does not follow the same pattern: the highly symmetrical neopentane packs well in the crystal and melts at 256.4 K, far higher than pentane (143.3 K).

Extra depth. This principle also explains why petrol with highly branched alkanes (high octane number, e.g. 2,2,4-trimethylpentane) burns more smoothly: branched alkanes are more resistant to premature ignition (knocking), although the octane scale itself is not in the NCERT text. In the case of alkenes and alkyl halides the same branching effect on boiling points is seen.

Conclusion: The boiling point decreases with increasing branching, because the more spherical molecule has a smaller surface area and hence weaker van der Waals forces.

Concept. Two different mechanisms operate. Without peroxide, HBr adds by electrophilic addition (ionic mechanism) and follows Markovnikov’s rule. With benzoyl peroxide, the same reagent adds by a free-radical chain mechanism and the orientation is reversed: anti-Markovnikov addition (peroxide or Kharasch effect, observed by M.S. Kharasch and F.R. Mayo in 1933).

A. Without peroxide: electrophilic addition (Markovnikov)

1. Attack of the electrophile. The π bond of propene attacks H⁺ of H–Br. Two carbocations are possible: (a) CH₃–CH₂–CH₂⁺ (primary, 1°) by adding H to C-2, or (b) CH₃–CH⁺–CH₃ (secondary, 2°) by adding H to C-1.
2. Stability decides. The 2° carbocation (b) is stabilised by two electron-releasing methyl groups (+I effect and hyperconjugation) and is more stable than the 1° ion (a), so it is formed faster and predominates.
3. Attack of the nucleophile. Br⁻ attacks the 2° carbocation: CH₃–CH⁺–CH₃ + Br⁻ → CH₃–CHBr–CH₃ (2-bromopropane, major product).

B. With benzoyl peroxide: free-radical chain (anti-Markovnikov)

In step (iii) Br• could add at C-1 (giving the more stable secondary radical CH₃–CH•–CH₂Br) or at C-2 (giving the less stable primary radical •CH₂–CHBr–CH₃). The more stable secondary radical is formed, so bromine ends up on C-1 and the product is 1-bromopropane. Radical stability order: 3° > 2° > 1°.

Why only HBr? With HCl the H–Cl bond (430.5 kJ mol⁻¹) is too strong for the radical to cleave in step (iv). With HI the H–I bond (296.8 kJ mol⁻¹) is cleaved easily but iodine radicals combine with each other to form I₂ instead of adding to the double bond. Only the H–Br bond (363.7 kJ mol⁻¹) is just right, so the peroxide effect is shown only by HBr.

Extra depth. The same dichotomy applies to every unsymmetrical alkene: with HBr alone, but-1-ene gives 2-bromobutane; with peroxide, 1-bromobutane. Markovnikov addition of HBr is a two-step ionic process whose rate-determining step is the carbocation formation; the peroxide route is a chain process in which a few peroxide molecules are enough to convert many alkene molecules. Always store HBr/alkene reaction mixtures free from peroxides (old ethers, air) if the Markovnikov product is wanted.

Conclusion: Without peroxide the 2° carbocation gives 2-bromopropane (Markovnikov); with benzoyl peroxide the 2° radical gives 1-bromopropane (anti-Markovnikov).

Concept. o-Xylene can be written with two Kekulé structures that differ in the position of the double bonds relative to the two methyl groups. Ozonolysis cleaves every C=C, so the two structures predict different fragments. The experimental formation of all the fragments shows that both structures contribute, supporting the idea of oscillating double bonds in benzene.

1. Number the ring carbons so that the two CH₃ groups are on C1 and C2.
2. In structure I the C1=C2 double bond (carrying both CH₃ groups) is cleaved to two C=O groups; each CH₃–C(=O) unit is still bonded to the next carbon, which becomes CHO: CH₃–CO–CHO (twice). The third double bond, C4=C5, gives OHC–CHO.
3. In structure II the C1–C2 bond between the two methyl-bearing carbons is a single bond, so the two CH₃–C(=O) units stay joined: CH₃–CO–CO–CH₃; the other two double bonds give two OHC–CHO.
4. If benzene had only one fixed Kekulé structure, o-xylene would give only one of the two product sets above.

Inference. The three products (methylglyoxal, glyoxal, butane-2,3-dione) are all obtained from o-xylene, which means o-xylene behaves as a mixture of, or hybrid of, structures I and II. Hence the double bonds in benzene are not fixed but delocalised/oscillating — this supports Kekulé’s modified proposal and the resonance picture, with the result that benzene gives only one 1,2-disubstituted product.

Extra depth. The ozonolysis of benzene itself gives three molecules of glyoxal (a triozonide), in agreement with three double bonds. The fact that o-xylene gives both sets of fragments shows that the two Kekulé forms are equally important; resonance theory goes one step further and says that the real molecule is neither form but a hybrid with equal C–C bonds (139 pm).

Conclusion: Ozonolysis of o-xylene gives methylglyoxal, glyoxal and butane-2,3-dione, which is explained only if both Kekulé structures (oscillating double bonds) are present.

Concept. Acidic behaviour of a hydrocarbon is its ability to lose the hydrogen attached to carbon as H⁺. It depends on the hybridisation of the carbon carrying that hydrogen: the greater the s-character of the C–H bond orbital, the more electronegative the carbon, and the more easily the C–H pair is held by carbon, leaving the proton free to leave and the carbanion more stable.

Order of decreasing acidic behaviour: ethyne > benzene > n-hexane.

1. In ethyne the H atoms are on sp-hybridised carbons (50% s-character). An s orbital is closer to the nucleus than a p orbital, so electrons in an sp orbital are held more tightly: the sp carbon is the most electronegative.
2. This carbon attracts the shared C–H electron pair strongly, so the hydrogen can leave as a proton relatively easily: HC≡CH + Na → HC≡C⁻Na⁺ + ½H₂ (and with NaNH₂).
3. In benzene the carbons are sp² (33% s), in n-hexane sp³ (25% s), so they hold the C–H pair less strongly; hexane is practically non-acidic, benzene only slightly more.

Experimental support. Only ethyne (and other terminal alkynes) reacts with sodium metal or sodamide to liberate H₂/NH₃; benzene and hexane do not.

Extra depth. The conjugate base of ethyne, the ethynide ion HC≡C⁻, has its lone pair in an sp orbital (close to the nucleus, hence stable); the conjugate base of hexane would have the lone pair in an sp³ orbital (further from the nucleus, hence unstable). The same argument gives the order HC≡CH > H₂C=CH₂ > CH₃–CH₃ and shows that benzene, though not very acidic, is a stronger acid than an alkane. The pK_a values (about 25 for ethyne, 43 for benzene, 50 for alkanes) follow the same sequence (supplementary data).

Conclusion: Decreasing acidic behaviour: ethyne > benzene > n-hexane, because the s-character (and hence the electronegativity) of the carbon bearing hydrogen is 50% (sp), 33% (sp²) and 25% (sp³) respectively.

Concept. Benzene has a cloud of six delocalised π electrons above and below the plane of the ring. This electron-rich cloud attracts electron-seeking reagents (electrophiles) and repels electron-rich reagents (nucleophiles).

1. Electron-rich ring. The π clouds above and below the ring make the ring a source of electrons, just like the π bond of an alkene, so an electrophile E⁺ (NO₂⁺, Cl⁺, R⁺, RCO⁺) is attracted to it.
2. Nucleophiles are repelled. A nucleophile (OH⁻, CN⁻, NH₂⁻) carries a lone pair and a negative charge; it is repelled by the electron cloud and cannot approach the ring easily.
3. Stable intermediate for electrophilic attack. The attack by E⁺ gives the arenium ion (σ-complex) whose positive charge is spread over three carbons by resonance; it then loses H⁺ and restores the aromatic ring. The whole sequence preserves aromaticity (substitution, not addition).
4. Unstable intermediate for nucleophilic attack. A nucleophile adding to the ring would give a carbanion with the extra negative charge on a ring carbon; an unsubstituted benzene ring has nothing to stabilise this charge, so the intermediate is of high energy and aromaticity is lost.
5. Leaving group. Benzene has no good leaving group (H⁻ is a very poor one), so even if the nucleophile added, substitution could not complete.

Extra depth. Nucleophilic substitution in benzene derivatives becomes possible when strongly electron-withdrawing groups are present at ortho/para positions to a leaving group (for example 2,4-dinitrochlorobenzene reacts with hydroxide): the nitro groups stabilise the carbanion intermediate by resonance. In plain benzene such stabilisation is absent. You will study this in Class XII (haloarenes).

Conclusion: Because of its delocalised π electron cloud, benzene is electron-rich: it attracts electrophiles (easy S_E) and repels nucleophiles (difficult S_N).

Concept. Benzene can be made from alkynes by cyclic polymerisation and from alkanes by aromatisation (reforming). Alkenes first have to be converted into the alkyne.

(i) Ethyne to benzene: cyclic polymerisation

Three molecules of ethyne trimerise in a red-hot iron tube at 873 K to give benzene: 3C₂H₂ → C₆H₆ (eq. 9.69). This is the best route from aliphatic to aromatic compounds.

(ii) Ethene to benzene: ethene → ethyne → benzene

1. Addition of bromine: CH₂=CH₂ + Br₂ → BrCH₂–CH₂Br (1,2-dibromoethane).
2. Dehydrobromination with alcoholic KOH: BrCH₂CH₂Br → CH₂=CHBr + HBr (bromoethene).
3. Second dehydrobromination with sodamide: CH₂=CHBr + NaNH₂ → HC≡CH + NaBr + NH₃.
4. Cyclic polymerisation of ethyne as in (i).

(iii) Hexane to benzene: aromatisation (reforming)

n-Alkanes having six or more carbon atoms are dehydrogenated and cyclised when heated to 773 K at 10–20 atmospheric pressure in the presence of oxides of vanadium, molybdenum or chromium supported over alumina. Hexane loses four molecules of H₂ (cyclisation + two further dehydrogenations) to give benzene.

Extra depth. The same aromatisation method converts n-heptane to toluene and is the basis of ‘catalytic reforming’ used in refineries to raise the octane number of petrol. The route in (ii) is a 4-step sequence; an alternative first step is dehydrohalogenation of the dihalide with alcoholic KOH followed by sodamide, which is the one given in the NCERT text for preparing alkynes from vicinal dihalides.

Conclusion: (i) Pass ethyne through a red-hot iron tube at 873 K; (ii) CH₂=CH₂ → BrCH₂CH₂Br → CH₂=CHBr → HC≡CH → benzene; (iii) heat hexane at 773 K, 10–20 atm over Cr₂O₃/V₂O₅/Mo₂O₃ on alumina.

Concept. Hydrogenation (H₂, Ni/Pd/Pt) adds H₂ across C=C without changing the carbon skeleton. So the required alkenes must have the carbon skeleton of 2-methylbutane, CH₃–CH(CH₃)–CH₂–CH₃, with a C=C inserted at every position where two adjacent carbons each have at least one hydrogen to lose.

1. Write the skeleton and number it: CH₃–CH(CH₃)–CH₂–CH₃ with the chain carbons numbered C1 to C4 and the methyl branch on C-2.
2. List all pairs of adjacent carbons that can lose one H each: C1–C2, C2–C3, C3–C4 and C2–(branch CH₃).
3. C1–C2 and C2–branch give the same compound (the two CH₃ groups on C-2 are identical), so there are three different alkenes.

All three on hydrogenation: CH₂=C(CH₃)CH₂CH₃, (CH₃)₂C=CHCH₃ and (CH₃)₂CHCH=CH₂ each + H₂ Ni⟶ CH₃–CH(CH₃)–CH₂–CH₃ (2-methylbutane, isopentane).

Extra depth. The same three alkenes are the only C₅H₁₀ alkenes with a branched skeleton. The straight-chain alkenes pent-1-ene and pent-2-ene give n-pentane (cis- and trans-pent-2-ene give the same alkane). Alkenes with the 2-methylbutane skeleton are structural isomers (position and chain relation), and 2-methylbut-2-ene is the most stable of the three because its C=C carries three alkyl groups (highest substitution, Saytzeff product).

Conclusion: 2-Methylbut-1-ene, 2-methylbut-2-ene and 3-methylbut-1-ene give 2-methylbutane on hydrogenation.

Concept. Reactivity towards an electrophile depends on the electron density of the ring: electron-releasing groups (+I, +R) increase it (activate), electron-withdrawing groups (–I, –R: –NO₂, halogens) decrease it (deactivate). The more deactivating groups present, the lower the reactivity.

(a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene

1. Chlorobenzene has only the weakly deactivating Cl: highest electron density of the three.
2. p-Nitrochlorobenzene carries Cl plus one strongly deactivating –NO₂: less reactive.
3. 2,4-Dinitrochlorobenzene carries Cl plus two –NO₂ groups: the ring is most electron-poor and least reactive.

(b) Toluene, p-H₃C–C₆H₄–NO₂, p-O₂N–C₆H₄–NO₂

1. Toluene has an electron-releasing CH₃: activated, the most reactive.
2. p-Nitrotoluene has one activating (CH₃) and one strongly deactivating (NO₂) group; the NO₂ effect dominates, so it is less reactive than toluene but more reactive than the dinitro compound because CH₃ partly compensates.
3. p-Dinitrobenzene has two strongly deactivating NO₂ groups and no activating group: least reactive.

Extra depth. This is the ring-reactivity ladder used in Friedel–Crafts and nitration problems: activated (–OH, –NH₂, –CH₃) > benzene > halogenobenzenes > nitro compounds. Strongly deactivated rings such as dinitro compounds do not undergo Friedel–Crafts reactions at all. Note the contrast with nucleophilic reactivity: 2,4-dinitrochlorobenzene is the most reactive of these towards nucleophiles (like OH⁻) because the nitro groups stabilise the carbanion intermediate — the order is reversed.

Conclusion: (a) Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene. (b) Toluene > p-nitrotoluene > p-dinitrobenzene.

Concept. Nitration is an electrophilic substitution in which the electrophile is the nitronium ion NO₂⁺. The easier the ring can donate electrons to NO₂⁺, the faster nitration occurs. So rank the three rings by electron density.

1. The methyl group of toluene pushes electrons into the ring (+I effect and hyperconjugation), especially at the ortho and para positions. The more electron-rich ring attacks NO₂⁺ faster, and the arenium ion formed is stabilised by the methyl group.
2. Benzene has no activating or deactivating group; its reactivity is the standard.
3. The two nitro groups of m-dinitrobenzene withdraw electrons strongly from the ring (–I and –R effects), so the ring is poor in electrons; the incoming NO₂⁺ is attacked with difficulty and the arenium ion is destabilised.

Practical evidence. Benzene is nitrated with the nitrating mixture (conc. HNO₃ + conc. H₂SO₄) on warming; toluene is nitrated under milder conditions; nitrobenzene needs more vigorous conditions (fuming acid and higher temperature) to give m-dinitrobenzene, and a third nitro group goes in with great difficulty.

Extra depth. Toluene gives mainly a mixture of o- and p-nitrotoluene (CH₃ is an ortho/para director), whereas the nitration of m-dinitrobenzene, if forced, gives 1,3,5-trinitrobenzene (the new NO₂ enters meta to both). The ranking toluene > benzene > nitro compounds is the same as in Ex 9.22 and is the basis of the rule that Friedel–Crafts alkylation cannot be done on a ring bearing a nitro group.

Conclusion: Toluene is nitrated most easily because the electron-releasing CH₃ group increases the electron density of the ring, while the two nitro groups of m-dinitrobenzene decrease it strongly; benzene lies in between.

Concept. Ethylation of benzene is a Friedel–Crafts alkylation: benzene reacts with an ethyl halide in the presence of a Lewis acid, which generates the electrophile C₂H₅⁺. Any strong Lewis acid (electron-pair acceptor) that is anhydrous can do this job.

1. Generation of the electrophile: C₂H₅–Cl + FeCl₃ → C₂H₅⁺ + [FeCl₄]⁻ (the Lewis acid accepts a lone pair from Cl).
2. Attack: benzene + C₂H₅⁺ → σ-complex (arenium ion).
3. Loss of proton: [FeCl₄]⁻ removes H⁺, giving ethylbenzene + HCl + FeCl₃ (the catalyst is regenerated).

Why anhydrous? Water reacts with the Lewis acid (AlCl₃, FeCl₃, BF₃) to form hydrated, inactive species and HCl, destroying the catalyst.

Extra depth. Ethylbenzene is the starting material for styrene (C₆H₅CH=CH₂), from which polystyrene is made. Other ethylating systems exist: ethene with AlCl₃/HCl or ethanol with BF₃ also generate the C₂H₅⁺ equivalent. A weak Lewis acid is enough for ethylation but ethyl cation does not rearrange, unlike n-propyl (see the isopropylbenzene question in the text).

Conclusion: Anhydrous ferric chloride (FeCl₃), a Lewis acid, can replace AlCl₃ in the ethylation of benzene (BF₃, SnCl₄ and ZnCl₂ can also be used).

Concept. The Wurtz reaction couples two alkyl groups: 2R–X + 2Na → R–R + 2NaX (dry ether). With identical halides it gives a single alkane containing twice the number of carbons of R, which is always an even number. To obtain an odd-numbered alkane, two different alkyl halides (R–X and R′–X) must be used, and then a mixture of three alkanes is obtained.

Example: propane (3 carbons) from CH₃Br and C₂H₅Br

1. Methyl bromide and ethyl bromide are mixed with sodium in dry ether.
2. The three kinds of coupling are equally possible, so ethane, propane and butane are all formed.
3. The mixture has a low yield of propane (the odd-carbon alkane), and the alkanes have similar properties; separating them by fractional distillation is difficult and wasteful.

Better routes to odd-numbered alkanes: decarboxylation of the sodium salt of a carboxylic acid with soda lime gives an alkane with one carbon less (e.g. sodium butanoate → propane, Problem 9.6); catalytic hydrogenation of an alkene or alkyne (propene or propyne → propane); reduction of an alkyl halide with Zn/H⁺ (1-chloropropane → propane).

Extra depth. Wurtz reaction also fails for tertiary alkyl halides (they undergo elimination to alkenes instead of coupling) and is not suitable for preparing methane (a coupling cannot give a single carbon). It is best for symmetrical alkanes such as ethane (from CH₃Br), butane (from C₂H₅Br) or 2,3-dimethylbutane (from 2-bromopropane).

Conclusion: Wurtz reaction with two different alkyl halides gives a mixture of three alkanes (e.g. ethane, propane, butane), so it is not used for odd-carbon alkanes; it is used only for even-carbon symmetrical alkanes.